For the reaction at 298 K,2 A^{+}B rightarr | vedclass

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(B) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative. $\Delta G = \Delta H - T\Delta S < 0$. At equilibrium,$

Vedclass · Accepted Answer

$2000$

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(B) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative. $\Delta G = \Delta H - T\Delta S At equilibrium,$\Delta G = 0$,so $T = \frac{\Delta H}{\Delta S}$. Given $\Delta H = 400 \ kJ \ mol^{-1}$ and $\Delta S = 0.2 \ kJ \ mol^{-1} \ K^{-1}$. $T = \frac{400}{0.2} = 2000 \ K$. Since $\Delta H$ and $\Delta S$ are both positive,the reaction becomes spontaneous at temperatures higher than the equilibrium temperature. Therefore,the reaction will become spontaneous above $2000 \ K$.

For the reaction at $298 \ K$,$2 \ A^{+}B \rightarrow C$. $\Delta H = 400 \ kJ \ mol^{-1}$ and $\Delta S = 0.2 \ kJ \ mol^{-1} \ K^{-1}$. The reaction will become spontaneous above $...... \ K$.

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